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The addition problem above shows four of the 24 different in [#permalink]
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Updated on: 06 Nov 2012, 03:17
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1,234 1,243 1,324 ..... .... +4,321 The addition problem above shows four of the 24 different integers that can be formed by using each of the digits 1,2,3,4 exact;y once in each integer. What is the sum of these 24 integers? A. 24,000 B. 26,664 C. 40,440 D. 60,000 E. 66,660
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Originally posted by student26 on 03 Nov 2010, 00:34.
Last edited by Bunuel on 06 Nov 2012, 03:17, edited 1 time in total.
Renamed the topic and edited the question.




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Re: The addition problem [#permalink]
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08 Feb 2011, 06:48




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03 Nov 2010, 00:45
student26 wrote: 1,234 1,243 1,324 ..... .... +4,321
The addition problem above shows four of the 24 different integers that can be formed by using each of the digits 1,2,3,4 exact;y once in each integer. What is the sum of these 24 integers?
A.24,000 B.26,664 C.40,440 D.60,000 E.66,660 Using the symmetry in the numbers involved (All formed using all possible combinations of 1,2,3,4), and we know there are 24 of them. We know there will be 6 each with the units digits as 1, as 2, as 3 and as 4. And the same holds true of the tens, hundreds and thousands digit. The sum is therefore = (1 + 10 + 100 + 1000) * (1*6 +2*6 +3*6 +4*6) = 1111 * 6 * 10 = 66660 Answer : e
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Re: The addition problem [#permalink]
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08 Feb 2011, 06:26
1,2,3,4 can be arranged in 4! = 24 ways The units place of all the integers will have six 1's, six 2's, six 3's and six 4's Likewise, The tens place of all the integers will have six 1's, six 2's, six 3's and six 4's The hundreds place of all the integers will have six 1's, six 2's, six 3's and six 4's The thousands place of all the integers will have six 1's, six 2's, six 3's and six 4's Addition always start from right(UNITS) to left(THOUSANDS); Units place addition; 6(1+2+3+4) = 60. Unit place of the result: 0 carried over to tens place: 6 Tens place addition; 6(1+2+3+4) = 60 + 6(Carried over from Units place) = 66 Tens place of the result: 6 carried over to hunderes place: 6 Hundreds place addition; 6(1+2+3+4) = 60 + 6(Carried over from tens place) = 66 Hundreds place of the result: 6 carried over to thousands place: 6 Thousands place addition; 6(1+2+3+4) = 60 + 6(Carried over from hundreds place) = 66 Thousands place of the result: 6 carried over to ten thousands place: 6 Ten thousands place of the result: 0+6(Carried over from thousands place) = 6 Result: 66660 Ans: "E"
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Re: The addition problem [#permalink]
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22 Jan 2013, 10:44
Bunuel wrote: Merging similar topics. Formulas for such kind of problems (just in case): 1. Sum of all the numbers which can be formed by using the \(n\) digits without repetition is: \((n1)!*(sum \ of \ the \ digits)*(111... \ n \ times)\). 2. Sum of all the numbers which can be formed by using the \(n\) digits ( repetition being allowed) is: \(n^{n1}*(sum \ of \ the \ digits)*(111... \ n \ times)\). Similar questions: nicequestionandagoodwaytosolve103523.htmlcansomeonehelp94836.htmlsumofall3digitnoswith88864.htmlpermutation88357.htmlsumof3digits78143.htmlCould you tell me the way to calculate the sum which the repetition is allowed? For example: from 1,2,3,4. how can we calculate the sum of four digit number that formed from 1,2,3,4 and repetition is allowed?
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Re: The addition problem [#permalink]
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22 Jan 2013, 21:07
hellscream wrote: Could you tell me the way to calculate the sum which the repetition is allowed? For example: from 1,2,3,4. how can we calculate the sum of four digit number that formed from 1,2,3,4 and repetition is allowed?
The logic is no different from 'no repetition allowed' question. The only thing different is the number of numbers you can make. How many numbers can you make using the four digits 1, 2, 3 and 4 if repetition is allowed? You can make 4*4*4*4 = 256 numbers (there are 4 options for each digit) 1111 1112 1121 ... and so on till 4444 By symmetry, each digit will appear equally in each place i.e. in unit's place, of the 256 numbers, 64 will have 1, 64 will have 2, 64 will have 3 and 64 will have 4. Same for 10s, 100s and 1000s place. Sum = 1000*(64*1 + 64*2 + 64*3 + 64*4) + 100*(64*1 + 64*2 + 64*3 + 64*4) + 10*(64*1 + 64*2 + 64*3 + 64*4) + 1*(64*1 + 64*2 + 64*3 + 64*4) = (1000 + 100 + 10 + 1)(64*1 + 64*2 + 64*3 + 64*4) = 1111*64*10 = 711040 or use the formula given by Bunuel above: Sum of all the numbers which can be formed by using the digits (repetition being allowed) is:\(n^{n1}\)*Sum of digits*(111...n times) =\(4^3*(1+2+3+4)*(1111) = 711040\) (Same calculation as above)
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The addition problem above shows four of the 24 different intege [#permalink]
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21 May 2013, 07:35
This can be solved much easier by realizing that, since the number of four term permutations is 4!, and that summing the a sequence to its reverse gives
1234 +4321 = 5555
1243 +3421 = 5555
we may see that there are 4!/2 pairings we can make, giving us
5555(12) = 66660



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Re: The addition problem above shows four of the 24 different in [#permalink]
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28 Sep 2014, 10:41
For those who could not memorize the formular, you can guess the answer in 30 secs: Since we have 24 numbers, we will have 6 of 1 thousand something, 6 of 2 thousand something, 6 of 3 thousand something, and 6 of 4 thousand something So, 6x1(thousand something) = 6 (thousand something) 6x2(thousand something) = 12 (thousand something) 6x3(thousand something) = 18 (thousand something) 6x4(thousand something) = 24 (thousand something) Add them all 6+12 +18 + 24 = 60 (thousand something) > E



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Re: The addition problem above shows four of the 24 different in [#permalink]
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15 Aug 2016, 14:38
each term is repeating 6 times..(total 24/4=4) now at unit...each term will repeat 6 times... (1x6+ 2x6 + 3x6 + 4x6 = 60) , so unit digit is "0" and 6 remaining.
repeating same.....total of 24 ten digit will be 60 + 6 from total of unit ,so ten digit will be 6.
only E has 60 as last two digits.



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Re: The addition problem above shows four of the 24 different in [#permalink]
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14 Sep 2017, 07:19
student26 wrote: 1,234 1,243 1,324 ..... .... +4,321
The addition problem above shows four of the 24 different integers that can be formed by using each of the digits 1,2,3,4 exact;y once in each integer. What is the sum of these 24 integers?
A. 24,000 B. 26,664 C. 40,440 D. 60,000 E. 66,660 Each digit will come at the respective place i.e units,tens, hundreds , thousands So calculate sum of each digit for the all the places for 4=4000+400+40+4 3=3000+300+30+3 2=2000+200+20+2 1=1000+100+10+1 Now calculate the sum of the sums of these digits =11110 Now we know that each digit is used 6 times therefore we have to multiply with 6 6*11110=66660 Hence E is our answer .
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Re: The addition problem above shows four of the 24 different in [#permalink]
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02 Oct 2017, 14:41
VeritasPrepKarishma wrote: hellscream wrote: Could you tell me the way to calculate the sum which the repetition is allowed? For example: from 1,2,3,4. how can we calculate the sum of four digit number that formed from 1,2,3,4 and repetition is allowed?
The logic is no different from 'no repetition allowed' question. The only thing different is the number of numbers you can make. How many numbers can you make using the four digits 1, 2, 3 and 4 if repetition is allowed? You can make 4*4*4*4 = 256 numbers (there are 4 options for each digit) 1111 1112 1121 ... and so on till 4444 By symmetry, each digit will appear equally in each place i.e. in unit's place, of the 256 numbers, 64 will have 1, 64 will have 2, 64 will have 3 and 64 will have 4. Same for 10s, 100s and 1000s place. Sum = 1000*(64*1 + 64*2 + 64*3 + 64*4) + 100*(64*1 + 64*2 + 64*3 + 64*4) + 10*(64*1 + 64*2 + 64*3 + 64*4) + 1*(64*1 + 64*2 + 64*3 + 64*4) = (1000 + 100 + 10 + 1)(64*1 + 64*2 + 64*3 + 64*4) = 1111*64*10 = 711040 or use the formula given by Bunuel above: Sum of all the numbers which can be formed by using the digits (repetition being allowed) is:\(n^{n1}\)*Sum of digits*(111...n times) =\(4^3*(1+2+3+4)*(1111) = 711040\) (Same calculation as above) How about using the Average*no. of terms formula where the average is found by adding the smallest number and the largest number divided by 2. It worked for this question. My question is will it work for all such types of questions?
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Re: The addition problem above shows four of the 24 different in [#permalink]
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02 Oct 2017, 20:57
sam2016 wrote: VeritasPrepKarishma wrote: hellscream wrote: Could you tell me the way to calculate the sum which the repetition is allowed? For example: from 1,2,3,4. how can we calculate the sum of four digit number that formed from 1,2,3,4 and repetition is allowed?
The logic is no different from 'no repetition allowed' question. The only thing different is the number of numbers you can make. How many numbers can you make using the four digits 1, 2, 3 and 4 if repetition is allowed? You can make 4*4*4*4 = 256 numbers (there are 4 options for each digit) 1111 1112 1121 ... and so on till 4444 By symmetry, each digit will appear equally in each place i.e. in unit's place, of the 256 numbers, 64 will have 1, 64 will have 2, 64 will have 3 and 64 will have 4. Same for 10s, 100s and 1000s place. Sum = 1000*(64*1 + 64*2 + 64*3 + 64*4) + 100*(64*1 + 64*2 + 64*3 + 64*4) + 10*(64*1 + 64*2 + 64*3 + 64*4) + 1*(64*1 + 64*2 + 64*3 + 64*4) = (1000 + 100 + 10 + 1)(64*1 + 64*2 + 64*3 + 64*4) = 1111*64*10 = 711040 or use the formula given by Bunuel above: Sum of all the numbers which can be formed by using the digits (repetition being allowed) is:\(n^{n1}\)*Sum of digits*(111...n times) =\(4^3*(1+2+3+4)*(1111) = 711040\) (Same calculation as above) How about using the Average*no. of terms formula where the average is found by adding the smallest number and the largest number divided by 2. It worked for this question. My question is will it work for all such types of questions? Sum of terms = Average * Number of terms holds for all sets of numbers Average = (First term + Last term)/2 holds for an Arithmetic Progression only
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Re: The addition problem above shows four of the 24 different in [#permalink]
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03 Oct 2017, 12:30
VeritasPrepKarishma wrote: sam2016 wrote: How about using the Average*no. of terms formula where the average is found by adding the smallest number and the largest number divided by 2. It worked for this question. My question is will it work for all such types of questions?
Sum of terms = Average * Number of terms holds for all sets of numbers Average = (First term + Last term)/2 holds for an Arithmetic Progression only Yes, you are right. But what was the arithmetic progression in this question? And do such types of questions always have arithmetic progression so that I can use the abovementioned formula for such questions?
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Re: The addition problem above shows four of the 24 different in [#permalink]
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03 Oct 2017, 21:45
sam2016 wrote: VeritasPrepKarishma wrote: sam2016 wrote: How about using the Average*no. of terms formula where the average is found by adding the smallest number and the largest number divided by 2. It worked for this question. My question is will it work for all such types of questions?
Sum of terms = Average * Number of terms holds for all sets of numbers Average = (First term + Last term)/2 holds for an Arithmetic Progression only Yes, you are right. But what was the arithmetic progression in this question? And do such types of questions always have arithmetic progression so that I can use the abovementioned formula for such questions? There isn't and hence, I haven't used this formula. Note that Avg concept will not work when the digits are say 1, 2, 4, 6. It works in this case because of the symmetry of the digits 1, 2, 3 and 4.
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Re: The addition problem above shows four of the 24 different in [#permalink]
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28 Mar 2018, 07:53
We have 6 numbers with 1XXX, 6 with 2XXX 6 with 3XXX and 6 with 4XXX.
6*1000+6*2000+6*3000+6*4000= 60000. Hence, the solution must be higher than 60000. Only E fits.



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Re: The addition problem above shows four of the 24 different in [#permalink]
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11 Jul 2018, 06:01
What's the level of this question? +600?



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